Mathematics Thread

[ben1996123]

orite yeah i misunderstood the question

goes through (0,1), (0,-1), centre (0,k), find ecuación

Spoiler

necks queschun:

find the maximum and minimum curvature of x^n and the coordinates where they occur

 

[vcuber13]

what do you mean by max and min curvature? the sharpest curve?

 

[vcuber13]

is there anyway to evaluate a definite integral by hand, when its anti derivative is not elementary

 

[A Leman]

is there anyway to evaluate a definite integral by hand, when its anti derivative is not elementary

Numerical approximation can be an entire class. It depends upon how accurate you want to be. If you literally mean "by hand". then I would use Simpson's rule. It is quick and easy. If you have a specific problem you want evaluated, please share. There may be a better way for your case.

 

[ben1996123]

is there anyway to evaluate a definite integral by hand, when its anti derivative is not elementary

you can change from cartesian to polar coordinates sometimes which dose stuff sometimes. like \( \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}dx \). if you change to polar coordinates then it turns into something else wich you can integræt

what do you mean by max and min curvature? the sharpest curve?

curvature

Numerical approximation can be an entire class. It depends upon how accurate you want to be. If you literally mean "by hand". then I would use Simpson's rule. It is quick and easy. If you have a specific problem you want evaluated, please share. There may be a better way for your case.

ew guessing
or just type it into wulfram or just make a program to do it for you

okso ill do my question because wyenot

Spoiler

\( \kappa =\frac{\frac{d^2y}{dx^2}}{(1+\frac{dy}{dx})^{\frac{3}{2}}} \)

\( \kappa =\frac{n(n-1)x^{n-2}}{(1+(nx^{n-1})^2)^{\frac{3}{2}}} \)

\( \frac{d\kappa}{dx}=n(n-1)x^{n-5}\left ( \frac{(n-2)x^2}{(n^2x^{2n-2}+1)^{\frac{3}{2}}}-\frac{3(n-1)n^2x^{2n}}{(n^2x^{2n-2}+1)^{\frac{5}{2}}} \right )=0 \)

\( n(n-1)x^{n-5}=0 \)

\( x=0 \)

\( \frac{(n-2)x^2}{(n^2x^{2n-2}+1)^{\frac{3}{2}}}-\frac{3(n-1)n^2x^{2n}}{(n^2x^{2n-2}+1)^{\frac{5}{2}}}=0 \)

\( \frac{(n-2)x^2}{(n^2x^{2n-2}+1)^{\frac{3}{2}}}=\frac{3(n-1)n^2x^{2n}}{(n^2x^{2n-2}+1)^{\frac{5}{2}}} \)

\( \frac{(n-2)x^2(n^2x^{2n-2}+1)^{\frac{5}{2}}}{(n^2x^{2n-2}+1)^{\frac{3}{2}}}=3(n-1)n^2x^{2n} \)

\( (n-2)x^2(n^2x^{2n-2}+1)=3(n-1)n^2x^{2n} \)

\( n^2x^{2n}+x^2=\frac{3(n-1)n^2x^{2n}}{n-2} \)

\( n^2+x^{2-2n}=\frac{3(n-1)n^2}{n-2} \)

\( x^{2-2n}=\frac{3(n-1)n^2}{n-2}-n^2=\frac{2n^3-n^2}{n-2} \)

\( x=\left(\frac{n-2}{2n^3-n^2}\right)^{\frac{1}{2n-2}} \)

curvature at that point is \( \kappa=\frac{n(n-1)\left(\frac{n-2}{2n^3-n^2}\right)^{\frac{n-2}{2n-2}}}{\left(\frac{n-2}{2n-1}+1\right)^{\frac{3}{2}}} \)

next question

what is \( \int \frac{1}{1+sin(x)}dx \)

 

[vcuber13]

Spoiler

according to wolfram this is wrong
= int (1 - sinx)/(1-sin^2 x) dx
= int (1 - sinx)/(cos^2 x) dx <---- apparently this is not equal to the previous line
= int sec^2 x dx - int tanxsecx dx
= tanx - secx

 

[ben1996123]

Spoiler

according to wolfram this is wrong
= int (1 - sinx)/(1-sin^2 x) dx
= int (1 - sinx)/(cos^2 x) dx <---- apparently this is not equal to the previous line
= int sec^2 x dx - int tanxsecx dx
= tanx - secx

Spoiler

hmm...

true, false, true

what ? ? ?

edit: actually its because wulfram doesnt realise that tan²x+1=sec²x means that you can have "+1+c" in an integral, and you can redefine that as "+c" so it says false because there are no solutions to "stuf"="stuf+1"

yeah, its actually tanx-secx-1, and wulfram didnt realise that -1+c is a constant

Spoiler

anyway, this.

\( \int\frac{1}{1+sin(x)}dx \)

\( t=tan(\frac{x}{2}),sin(x)=\frac{2t}{t^2+1},dx=\frac{2}{t^2+1}dt \)

\( \int\frac{1}{1+sin(x)}dx=\int\frac{\frac{2}{t^2+1}}{1+\frac{2t}{t^2+1}}dt=\int\frac{2}{(t+1)^2}dt=\frac{-2}{t+1}+c=\frac{-2}{tan(\frac{x}{2})+1}+c \)

 

[ben1996123]

okso find an ecuación of a circlewave

 

[CarlBrannen]

Nice problem. Seems like a good target for a Fourier transform. Look for a solution in the form:

a_1 \sin(x) + a_3 \sin(3x) + a_5 \sin(5x) + ...

Make these fit between 0 and 2 pi by integration. Average of sin^2(kx) = 1/2 so get

a_n (2 \pi) (1/2) = \int_0^{\pi} \sin(nx) \sqrt(1-(x-\pi/2)^2/(\pi/2)^2 )dx + \int_\pi^{2\pi} \sin(nx) sqrt(1-(x-3\pi/2)^2/(\pi/2)^2 ) dx.

Solving for a_n gives a formal solution. The two integrals are obviously equal and shifting by pi/2 gives:

a_n = (2/\pi) \int_{-\pi/2}^{+\pi/2} \cos(nx) \sqrt(1-x^2/(\pi/2)^2 ) dx = (4/\pi)\int_0^{+\pi/2} \cos(nx) \sqrt(1-x^2/(\pi/2)^2 ) dx

= 2 \int_0^1 \cos(nx\pi/2) \sqrt(1-x^2) dx

My instinct is that looks like an integral that solves with complex analysis so it can be looked up in a table. Don't have any handy and I'm too lazy to work it out.

 

[ben1996123]

Nice problem. Seems like a good target for a Fourier transform. Look for a solution in the form:

yar or floor(stuf) or sgn(stuf)

I got y=sqrt(1/4-(x-floor(x)-1/2)^2)sgn(sin(pi x)) and arek got y=sqrt(1-((x-1)-2floor((x-1)/2)-1)^2)*(2floor((x+1)/2-2floor((x+1)/4))-1)

 

[CarlBrannen]

Here's another go.

What we're looking for is a modification to y = Cos(x) that changes the shape of y. So how about trying to use the form

y = f ( Cos (x) ) where
y^2 + (x-pi/2)^2 = (pi/2)^2, so
f(Cos(x)) = sqrt((pi/2)^2 - (x-pi/2)^2) = sqrt(x(pi-x))

Replacing x in the above with ArcCos(x) gives:
f(x) = sqrt(ArcCos(x)(Pi- ArcCos(x)) so

y = f(Cos(x)) = sqrt[ArcCos(Cos(x)) ( Pi - ArcCos(Cos(x))) ], which is correct for 0<= x <= pi, but has the wrong signs when Sin(x)<0. So to fix that, we end up with a thing that divides by zero when Sin(x) = 0, at which points we need to get an answer of 0.

y = Sin(x) sqrt[ArcCos(Cos(x)) (Pi - ArcCos(Cos(x)))/(1 - Cos^2(x))].

 

[A Leman]

Here's another go.

What we're looking for is a modification to y = Cos(x) that changes the shape of y. So how about trying to use the form

y = f ( Cos (x) ) where
y^2 + (x-pi/2)^2 = (pi/2)^2, so
f(Cos(x)) = sqrt((pi/2)^2 - (x-pi/2)^2) = sqrt(x(pi-x))

Replacing x in the above with ArcCos(x) gives:
f(x) = sqrt(ArcCos(x)(Pi- ArcCos(x)) so

y = f(Cos(x)) = sqrt[ArcCos(Cos(x)) ( Pi - ArcCos(Cos(x))) ], which is correct for 0<= x <= pi, but has the wrong signs when Sin(x)<0. So to fix that, we end up with a thing that divides by zero when Sin(x) = 0, at which points we need to get an answer of 0.

y = Sin(x) sqrt[ArcCos(Cos(x)) (Pi - ArcCos(Cos(x)))/(1 - Cos^2(x))].

This is very cool! I managed to get y=1/2*sgn(Sin(x))*sqrt(2ArcCos(Cos(x))*2ArcSin(Cos(x)+pi)) by rewriting your result.

 

[Ickathu]

Here's a problem that I'm stuck on:
Given m=2n+1, what is the inverse of 2 modulo m in terms of n?

Here's what I tried:
If m=2n+1, then 2n=m-1. Then, we get (2)(n) == -1 mod m. Multiplying each side by -1, we get: (2)(-n) == 1 mod m, which means that the inverse of 2 modulo m is -n, but that's not the right answer.

0 <= mk-n <= m-1
0 <= m-n <= m-1; n <= m and -n <= -1; n >= 1, so that would make m-n the right answer, but once again, it says this is wrong.

If I try k=2, I get:
0 <= 2m-n <= m-1
n <= 2m and m+1 <= n, or m+1 <= 2m, giving 1 <= m.

leaving k as k
0 <= mk-n <= m-1
n <= mk and mk-m+1 <= n, or mk-m+1 <= mk, giving 1 <= m

(2)(mk-n) == 1 mod m
(2)(mk-n) = mk + 1
2mk - 2n = mk+1
mk = 2n+1, which is the original equation, but it's no the answer either.


Also, this is for an online (AoPS) class, so, it might be the correct answer that I have, I could just have it formatted wrong, or something.

 

[MTGjumper]

Is it not just n+1?

2*(n+1) = 2n + 2 = 1 mod m

 

[Ickathu]

yes indeed it is. Thanks! I totally missed that haha

 

[Ickathu]

Find the number of integers 0 \le n \le 10000 that satisfy n^2 == 1 mod 10001
I've already got n=1 and n=10000, because ((b-1)^2 == 1 mod b) and (1^1 == 1 mod x), but are there more solutions? And how do I find them? I thought I'd have to do something the the prime factorization of 10001.
n^2 == 1 mod 10001
n^2 - 1 == 0 mod 10001
(n-1)(n+1) == 0 mod 10001, but I'm not sure what to do from here, since I can't divide and I don't know what the inverses are of (n-1) and (n+1)

 

[ben1996123]

Find the number of integers 0 \le n \le 10000 that satisfy n^2 == 1 mod 10001
I've already got n=1 and n=10000, because ((b-1)^2 == 1 mod b) and (1^1 == 1 mod x), but are there more solutions? And how do I find them? I thought I'd have to do something the the prime factorization of 10001.
n^2 == 1 mod 10001
n^2 - 1 == 0 mod 10001
(n-1)(n+1) == 0 mod 10001, but I'm not sure what to do from here, since I can't divide and I don't know what the inverses are of (n-1) and (n+1)

Spoiler: answers

1, 2191, 7810, 10000

 

[Ickathu]

but how? I can plug the congruence into wolfram if I just want the answers.

 

[Ickathu]

Nevermind, I've got it now. It was pretty easy once I realized what to do. That'd be a lot more casework if 10001 was divisible by more than 2 primes...

 

[Ickathu]

Find the number of distinct pairs of integers (x,y) such that 0<x<y and sqrt 1984 = sqrt x + sqrt y.

I tried subtracting sqrt y from each side, then squared the equation, but that just gave me 1984-2sqrt(1984y) + y = x, which I can't really work with.